A082040 a(n) = 9*n^2 + 3*n + 1.

1, 13, 43, 91, 157, 241, 343, 463, 601, 757, 931, 1123, 1333, 1561, 1807, 2071, 2353, 2653, 2971, 3307, 3661, 4033, 4423, 4831, 5257, 5701, 6163, 6643, 7141, 7657, 8191, 8743, 9313, 9901, 10507, 11131, 11773, 12433, 13111, 13807, 14521, 15253, 16003, 16771, 17557

Offset: 0

Author

Paul Barry, Apr 02 2003

Keywords

• nonn
• easy

Comments

4th row of A082039, case k = 3 of family T(n,k) = k^2*n^2 + k*n + 1.

a(n)^2 = 81*n^4 + 54*n^3 + 27*n^2 + 6*n + 1 = (24*((3*((3*n^2 + n)/2)^2 + ((3*n^2 + n)/2))/2) + 1). Therefore, (a(n)^2 - 1)/24 is a second pentagonal number (A005449) of index number equal to the n-th second pentagonal number. For example, a(30) = 8191 and (8191^2 - 1)/24 = (67092481 - 1)/24 = 2795520, the 1365th second pentagonal number. 1365 is the 30th second pentagonal number. - Raphie Frank, Sep 19 2012

For n >= 1, a(n) is the number of vertices in the hex derived network HDN1(n+1) from the Manuel et al. reference (see HFN1(4) in Fig. 8). - Emeric Deutsch, May 21 2018

4*a(n) - 3 is a square. - Muniru A Asiru, May 24 2018

Links

• Muniru A Asiru, Table of n, a(n) for n = 0..5000
• Nicolay Avilov, Graphic illustration of sequence members
• P. Manuel, R. Bharati, I. Rajasingh, and Chris Monica M, On minimum metric dimension of honeycomb networks, Journal of Discrete Algorithms, Vol. 6 (2008), pp. 20-27.
• Leo Tavares, Snowflake illustration.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A002061, A005449, A045945, A054569, A082039.

Partial sums of A019557.

Cf. A000290, A003215.

Programs

GAP

List([0..50], n->9*n^2+3*n+1); # Muniru A Asiru, May 21 2018

Maple

seq(9*n^2+3*n+1,n=0..50); # Muniru A Asiru, May 21 2018

PARI

a(n)=9*n^2+3*n+1 \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 18*n + a(n-1) - 6 with n > 0, a(0)=1. - Vincenzo Librandi, Aug 08 2010

a(n) = A045945(n) + 1: subsequence of A002061. - Muniru A Asiru, May 26 2018

a(n) = A003215(n) + 6*A000290(n). - Leo Tavares, Jul 14 2023

From Elmo R. Oliveira, Oct 23 2024: (Start)

G.f.: (1 + 10*x + 7*x^2)/(1 - x)^3.

E.g.f.: (1 + 12*x + 9*x^2)*exp(x).

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)